Question
If $$f\left( x \right) = \frac{{{p^2} - 1}}{{{p^2} + 1}}{x^3} - 3x + \log \,2$$ is a decreasing function of $$x$$ in $$R$$ then the set of possible values of $$p$$ ( independent of $$x$$ ) is :
A.
$$\left[ { - 1,\,1} \right]$$
B.
$$\left[ {1,\,\infty } \right)$$
C.
$$\left( { - \infty ,\, - 1} \right]$$
D.
none of these
Answer :
$$\left[ { - 1,\,1} \right]$$
Solution :
$$\eqalign{
& f'\left( x \right) = \frac{{{p^2} - 1}}{{{p^2} + 1}}.3{x^2} - 3 < 0 \cr
& \Rightarrow \frac{{{p^2} - 1}}{{{p^2} + 1}}{x^2} < 1 \cr
& \Rightarrow \left( {{p^2} - 1} \right){x^2} - \left( {{p^2} + 1} \right) < 0 \cr} $$
The inequation is satisfied for all $$x\, \in \,R$$ if $${p^2} - 1 \leqslant 0,{\text{ i}}{\text{.e}}{\text{., }} - 1 \leqslant p \leqslant 1$$